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Saturday, January 13, 2007

Computing the size of a convex hull

We know that in the comparison model, computing a planar convex hull takes Ω(n log n) time (or Ω(n log h, if you're picky) in any reasonable algebraic decision tree model. The proof (at least for n log n) is a simple reduction from sorting. There's a more involved proof by Yao that shows that even generating the set of vertices of the convex hull takes at least n log n time. (i.e ordering the vertices isn't the only hard part). Yao's proof is on the quadratic test model; tests on objects are allowed to be signs of quadratic polynomials, and such tests characterize all known convex hull algorithms. [Addendum: Ben-Or generalizes these results to higher-degree algebraic tests as well]

My question is: is it known whether estimating the size of the convex hull is also lower bounded by n log n (or n log h) in this model ? It seems like this should be true: I don't see an obvious way of using a size-estimation oracle to actually compute the hull (or its vertices) in linear time, but I don't know if there's a proof of this. Yao's proof doesn't appear to yield any direct insight into this question.

It's worth pointing out in the light of recent results by Chan, (and by Pătraşcu on point location), that we can solve the 2D convex hull problem in sub- n log n time; we pick our favorite machine model and sort the x-coordinates in sub-n log n time, and then run the Graham scan in linear time. Chan's paper achieves sub n log n bounds for the 3D convex hull and related problems.